If the magnetic field in a plane electromagnetic wave is given by B = 3 x 10^-8 sin(1.6 x 10^3 x + 4

Q: Why does $(kx + \omega t)$ mean the wave is traveling in the negative x-direction?

Think of a point on the wave with a constant phase. If $kx + \omega t = \text{constant}$, then as time $t$ increases, $x$ must decrease to keep the sum constant. Hence, the wave moves in the $-x$ direction.

Q: Why do we use $E_0 = cB_0$?

In a vacuum (or air), the ratio of the amplitudes of the electric and magnetic fields in an electromagnetic wave is always equal to the speed of light, $c$. This is a fundamental property derived from Maxwell's equations.

Q: How do I quickly remember the cross product rules for unit vectors?

Use the right-hand rule or the cyclic circle: $\mathbf{\hat{i}} \to \mathbf{\hat{j}} \to \mathbf{\hat{k}} \to \mathbf{\hat{i}}$. Going clockwise gives positive cross products (e.g., $\mathbf{\hat{i}} \times \mathbf{\hat{j}} = \mathbf{\hat{k}}$), and counter-clockwise gives negative (e.g., $\mathbf{\hat{k}} \times \mathbf{\hat{j}} = -\mathbf{\hat{i}}$).

Q: Can the phase angle be different for the electric and magnetic fields?

No, in a plane electromagnetic wave propagating in a vacuum or a linear isotropic dielectric, the electric and magnetic fields are always perfectly in phase. They reach their maxima and minima together.

Q: What if the medium wasn't a vacuum?

If the wave was in a medium with refractive index $n$, the speed of the wave would be $v = c/n$, and the relationship would become $E_0 = vB_0$. However, unless specified, we assume free space.

Question

If the magnetic field in a plane electromagnetic wave is given by $\mathbf{B} = 3 	imes 10^{-8} \sin(1.6 	imes 10^3 x + 48 	imes 10^{10} t) \mathbf{\hat{j}} 	ext{ T}$, then what will be expression for electric field?

Options:

(A) $\mathbf{E} = 60 \sin(1.6 	imes 10^3 x + 48 	imes 10^{10} t) \mathbf{\hat{k}} 	ext{ V/m}$

(B) $\mathbf{E} = 3 	imes 10^{-8} \sin(1.6 	imes 10^3 x + 48 	imes 10^{10} t) \mathbf{\hat{i}} 	ext{ V/m}$

(C) $\mathbf{E} = 3 	imes 10^{-8} \sin(1.6 	imes 10^3 x + 48 	imes 10^{10} t) \mathbf{\hat{j}} 	ext{ V/m}$

(D) $\mathbf{E} = 9 \sin(1.6 	imes 10^3 x + 48 	imes 10^{10} t) \mathbf{\hat{k}} 	ext{ V/m}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine an electromagnetic wave traveling through space. We are given the equation for its magnetic field, which is oscillating along the y-axis.

The equation is $\mathbf{B} = 3 	imes 10^{-8} \sin(1.6 	imes 10^3 x + 48 	imes 10^{10} t) \mathbf{\hat{j}} 	ext{ T}$.

Our goal is to construct the corresponding equation for the electric field. To do this, we need three pieces of information: the amplitude, the direction of propagation, and the direction of oscillation.

### The Master Equation for Amplitude

Let's start with the amplitude. In a vacuum, the amplitudes of the electric and magnetic fields are intimately connected by the speed of light.

This relationship is given by the master equation **$E_0 = cB_0$**.

By substituting the given magnetic field amplitude $B_0 = 3 	imes 10^{-8} 	ext{ T}$ and the speed of light $c = 3 	imes 10^8 	ext{ m/s}$, we can easily find the electric field amplitude.

$E_0 = (3 	imes 10^8) 	imes (3 	imes 10^{-8}) = 9 	ext{ V/m}$.

### Decoding the Phase for Propagation

Next, we need to figure out which way the wave is moving. The secret lies inside the sine function, specifically in the phase term $(1.6 	imes 10^3 x + 48 	imes 10^{10} t)$.

Notice that both the $x$ and $t$ terms have the same positive sign.

When the spatial and temporal coefficients share the same sign, it indicates that the wave is traveling in the **negative direction**.

Therefore, our wave is propagating along the negative x-axis, which means its velocity vector $\mathbf{\hat{v}}$ is in the **$-\mathbf{\hat{i}}$** direction.

### The Cross Product Rule

Now for the most crucial step: determining the direction of the electric field.

Electromagnetic waves are transverse, meaning the electric field, magnetic field, and the direction of propagation are all mutually perpendicular.

They follow a strict right-handed rule: **$\mathbf{\hat{E}} 	imes \mathbf{\hat{B}} = \mathbf{\hat{v}}$**.

We know the magnetic field is oscillating along $\mathbf{\hat{j}}$ and the wave is traveling along $-\mathbf{\hat{i}}$.

So, we must solve the cross product: $\mathbf{\hat{E}} 	imes \mathbf{\hat{j}} = -\mathbf{\hat{i}}$.

Using the cyclic properties of unit vectors, we know that $\mathbf{\hat{k}} 	imes \mathbf{\hat{j}} = -\mathbf{\hat{i}}$.

Thus, the electric field must be oscillating along the z-axis, in the **$\mathbf{\hat{k}}$** direction.

### Final Calculation

We now have all the puzzle pieces. The amplitude is $9 	ext{ V/m}$, the direction is $\mathbf{\hat{k}}$, and the phase remains identical to the magnetic field.

Putting it all together, we get our final expression.

$\mathbf{E} = 9 \sin(1.6 	imes 10^3 x + 48 	imes 10^{10} t) \mathbf{\hat{k}} 	ext{ V/m}$.

This perfectly matches option (d).

Animated Solution for Physics - Electromagnetic Waves: If the magnetic field in a plane electromagnetic wave is given by $B = 3 \times 1 0^{- 8} sin (1.6 \times 1 0^{3} x + 48 \times 1 0^{10} t) \hat{j} T$ , then what will be expression for electric field?

Select Answer:

Visualized Solution

Calculating $E_{0}$

Direction of Propagation

Direction of $E$

Final Equation

The Sigma Insight: Characteristics of Electromagnetic Waves

Analyzing the Setup

The Master Equation for Amplitude

Decoding the Phase for Propagation

The Cross Product Rule

Final Calculation

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